Question

If the refractive index of a material of equilateral prism is $\sqrt{3}$ , then angle of minimum deviation of the prism is

• A. 60${}^{\circ }$
• B. 45${}^{\circ }$
• C. 30${}^{\circ }$
• D. 75${}^{\circ }$

Answer 1

Answer: (A)

$A=60^{\circ },\mu =\sqrt{3},{\delta }_m=?$
$\mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\frac{A}{2}}\therefore {\delta }_m=60^{\circ }$